# The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

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• Volume of the balloon $V=\large\frac{4}{3}$$\pi r^3 Step 1: Volume of the balloon V=\large\frac{4}{3}$$\pi r^3$
Diff w.r.t $t$ we get,
$\large\frac{dv}{dt}=\frac{4}{3}$$\pi 3r^2\large\frac{dr}{dt} When t=0,r=3 units When t=3,r=6 units It is given \large\frac{dv}{dt}=constant=k(say) \therefore k=4\pi r^2\large\frac{dr}{dt} Separating the variables we get, \therefore kdt=4\pi r^2dr Step 2: Integrate on both sides k\int dt=4\pi\int r^2dr kt=4\pi \large\frac{r^3}{3}$$+c$------(1)
When $t=0,r=$3units
$\Rightarrow k(0)=\large\frac{4\pi(3)^3}{3}$$+c c=36\pi Step 3: When t=3sec,r=6units \Rightarrow k(3)=\large\frac{4\pi(6)^3}{3}$$+36\pi$
$3k=288\pi+36\pi$
$k=\large\frac{324\pi}{3}$
$\;\;\;=108\pi$
$\therefore$ Hence $c=36\pi$ and $k=108\pi$
Step 4:
Now substituting the values in equ(1)
$108\pi t=\large\frac{4}{3}$$\pi r^3+36\pi$
$4\pi r^3=36\pi-108\pi t$
$r^3=9-27t$
$\therefore r=\sqrt[\large 3]{9-27t}$
edited Nov 8, 2013